I am a French guest and I hope that my English isn't too bad...

So here is my issue: I consider an entire function $f$ which satisfies the following property for each complex number $z\in \mathbb{C}$:

$\forall ~ k \in \mathbb{N}^*$, there exists an entire function $G_k$ that satisfies $$f(z)=\exp_k(G_k(z))$$ where $\exp_k$ denotes $\exp \circ \exp \circ ...\circ \exp$, $k$-times.

In other words, I can take (as many times as I wish) the $\log$ of my function $f$, and it will always give an entire function that doesn't vanish on $\mathbb{C}$.

Is $f$ a constant? (surely different from $0$...)

Thanks everyone!

(PS: This forum is so cool!)

  • 5
    $\begingroup$ English is great! $\endgroup$ Feb 17, 2014 at 21:15
  • 1
    $\begingroup$ The remark "surely different of $0,1,\dots $" is misleading, and contributed to some wrong now-deleted answers. It's true that $f$ cannot be $0$, but the other constants are not excluded. For example, $f\equiv 1$ can be written as $\exp(2\pi i)$ and also as $\exp_2(\log(2\pi i))$ for an arbitrary choice of $\log$, and so on... $\endgroup$
    – user127096
    Mar 16, 2014 at 3:05

2 Answers 2


I must say that at first I was quite skeptical that such function could exist, but it turned out that one can prove its existence. I have written down the proof but I will give here only sketch it (you should not have any problems computing it for yourself, nice exercise)

Let us define $b_1=2\pi \rm i m_1$ and $\displaystyle b_n=2\pi \rm i m_n+ \log b_{n-1}$ and $$f_n(z)=\exp_{n}\left[b_n+ \frac{z}{b_1b_2\ldots b_{n-1}}\right],$$ where $\exp_{n}[z]=\exp\circ\cdots\circ\exp(z)$ $n$-times. One must find a sequence of integers $m_k>0$ such that the sequence $f_n(z)$ will converge on every compact set. With some more effort one can prove that limit function $g(z)=1+z+ \mathcal{O}(z^2)$ have the desired property.

  • $\begingroup$ I dont get it ? $\endgroup$
    – mick
    Oct 23, 2014 at 22:42
  • $\begingroup$ Try to do it as an exercise! Take $\{D_n\}$ compact exhaustion of $\mathbb{C}$ and show that $m_n$ can be taken so large that $|f_n-f_{n+1}|<1/n$ on D_n. By taking k times logarithm of $f_n$ and $f_{n+1}$ that they can be again $1/n$ apart on $D_n$ (increas $m_n$ if necessary) and see what happens. $\endgroup$
    – User113
    Oct 24, 2014 at 23:51

I think that the concept of 'order of an entire function' can help you to get an answer.
(See definition in W. Rudin : Exercise 2, Ch. 15 of ; or alternatively here : Wiki - Entire function )

Namely, your entire function $f$ is clearly of infinite order (well, at least for k>1), while constant functions -like polynomials- are entire functions of order zero ...

NB: of course I'm not considering the trivial case where the $G_k$ are constant (and non-zero), in which case $f$ would be constant as well ...

  • $\begingroup$ Yes, but the question here is: does such a (non-constant) function exist? Your answer definitely doesn't address the problem... $\endgroup$ Feb 19, 2014 at 6:06
  • $\begingroup$ The question was : is f constant ? $\endgroup$
    – MrA
    Feb 19, 2014 at 23:35
  • $\begingroup$ Daniel, not sure to understand your comment ... The initial question was : "is f constant ?" <br/>The elements given in my post imply : "No f cannot be constant" exception made of the trivial case mentioned in NB ... <br/>The main argument I provided is based on the order of an entire function. $\endgroup$
    – MrA
    Feb 19, 2014 at 23:47
  • $\begingroup$ Since there is a complex solution $z_0$ to $e^z=z$, then obviously the function $f(z)=z_0$ satisfies the condition, and is constant with $G_k(z)=z_0$. So there is certainly at least a constant function with the property of the question. The question then becomes: is there a non-constant function $f$ satisfying the condition, or are all such functions constant? $\endgroup$ Feb 20, 2014 at 19:47
  • $\begingroup$ @ Daniel Robert-Nicoud, sorry to insist but I don't get your logics here: the example you're providing in your last comment is just the 'trivial case' I was mentioning at the end of my answer, which I deemed not too informative for the problem at hand. (comment continued below) $\endgroup$
    – MrA
    Feb 20, 2014 at 22:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .